对于n对象，存在n!排列。考虑一组：

S = {a1, a2, a3 ..., an};

查找以上集合的置换的算法可以是：

foreach(item i : S) {
   /* all other item except i1 and i */
   foreach(item i1 : S - {i}) {
       foreach(item i2 : S - {i, i1}) {
          .
          .
          .
          foreach(item in : S - {i, i2, .... in-1}) {
             /* permutation list */
             P = { i1, i2, ...., in-1, in }; 
          }
       }
   }
}

显然，我们没有n for循环，但是可以递归构造此算法，直到获得nlist中的元素P。以下是使用上述算法进行置换的实际Java代码：

public static void 
permutations(Set<Integer> items, Stack<Integer> permutation, int size) {

    /* permutation stack has become equal to size that we require */
    if(permutation.size() == size) {
        /* print the permutation */
        System.out.println(Arrays.toString(permutation.toArray(new Integer[0])));
    }

    /* items available for permutation */
    Integer[] availableItems = items.toArray(new Integer[0]);
    for(Integer i : availableItems) {
        /* add current item */
        permutation.push(i);

        /* remove item from available item set */
        items.remove(i);

        /* pass it on for next permutation */
        permutations(items, permutation, size);

        /* pop and put the removed item back */
        items.add(permutation.pop());
    }
}

这是主要方法：

public static void main(String[] args) {
    // TODO Auto-generated method stub


    Set<Integer> s = new HashSet<Integer>();
    s.add(1);
    s.add(2);
    s.add(3);

    permutations(s, new Stack<Integer>(), s.size());
}

它打印了结果：

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]